What is x^2+y^2+9x-8y-62=0 ?

The solution to x^2+y^2+9x-8y-62=0 is Circle with (a,b)=(-9/2 ,4),r=(sqrt(393))/2

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x^2+y^2+9x-8y-62=0

x^{2} + y^{2} + 9 x - 8 y - 62 = 0

(a, b) = (- \frac{9}{2}, 4), r = \frac{393}{2}

Solution steps

x^{2} + y^{2} + 9 x - 8 y - 62 = 0

Rewrite

x^{2} + y^{2} + 9 x - 8 y - 62 = 0

in the form of the standard circle equation

(x - (- \frac{9}{2}))^{2} + (y - 4)^{2} = (\frac{393}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{9}{2}, 4), r = \frac{393}{2}

(x - 2)^{2} + (y + 2)^{2} = 2^{2}

x^{2} + y^{2} - 8 x + 8 y + 22 = 0

x^{2} + y^{2} + 3 x - 2 y - 1 = 0

x^{2} + y^{2} - 2 y - 31 = 0

x^{2} + y^{2} + 10 x - 14 y - 7 = 0

What is x^2+y^2+9x-8y-62=0 ?
The solution to x^2+y^2+9x-8y-62=0 is Circle with (a,b)=(-9/2 ,4),r=(sqrt(393))/2